Barrons AP Calculus

Figure  N6–10

SOLUTION: We know that if f ′ < 0 on an interval then f increases on the
interval, while if f ′> 0 then f decreases; also, if f ′ is increasing on an interval
then the graph of f is concave up on the interval, while if f ′ is decreasing then
the graph of f is concave down. These statements lead to the following
conclusions:

f   increases   on  [0,1]   and [3,5],  because f   ′   >   0   there;

but f   decreases   on  [1,3],  because f   ′   <   0   there;

also the    graph   of  f   is  concave down    on  [0,2],  because f   ′   is  decreasing;

but the graph   of  f   is  concave up  on  [2,5],  because f   ′is increasing.

Additionally, since f ′(1) = f ′(3) = 0, f has critical points at x = 1 and x = 3. As x
passes through 1, the sign of f ′ changes from positive to negative; as x passes
through 3, the sign of f ′ changes from negative to positive. Therefore f(1) is a
local maximum and f(3) a local minimum. Since f changes from concave down
to concave up at x = 2, there is an inflection point on the graph of f there.
These conclusions enable us to get the general shape of the curve, as displayed
in Figure N6–11a.